Question

Find the solutions to the following linear-quadratic systems algebraically. Select the ordered pair that is one of the correct solutions from among the choices below.

y = x² +3x + 8
y = 2x + 10

(-2, 6)
(0, 8)
(0,10)
(2, 14)

Answer 1

(-2, 6)

Explanation:

we have

`y=x^(2) +3x+8` ----> equation A

`y=2x+10` ----> equation B

equate equation A and equation B

`x^(2) +3x+8=2x+10`

`x^(2) +3x+8-2x-10=0`

`x^(2) +x-2=0`

The formula to solve a quadratic equation of the form

`ax^(2) +bx+c=0`

is equal to

`x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2) +x-2=0`

so

`a=1\\b=1\\c=-2`

substitute in the formula

`x=\frac{-1\pm\sqrt{1^(2)-4(1)(-2)}} {2(1)}`

`x=\frac{-1\pm√(9)} {2}`

`x=\frac{-1\pm3} {2}`

`x=\frac{-1+3} {2}=1`

`x=\frac{-1-3} {2}=-2`

Find the values of y

For x=1

`y=2(1)+10=12`

For x=-2

`y=2(-2)+10=6`

therefore

The solutions are (1,12) and (-2,6)